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alexira [117]
2 years ago
6

Which pair of samples contains the same number of oxygen atoms in each compound?0.20 mol Ba(OH)2 and 0.20 mol H2SO40.20 mol Br2O

and 0.20 mol HBrO0.10 mol Fe2O3 and 0.50 mol BaO0.10 mol Na2O and 0.10 mol Na2SO4
Chemistry
1 answer:
Whitepunk [10]2 years ago
4 0

Answer:

0.20 mol Br2O and 0.20 mol HBrO have the same number of oxygen atoms

Explanation:

<em>0.20 mol Ba(OH)2 and 0.20 mol H2SO4</em>

In Ba(OH)2 there are 2 moles of O atoms in every mol of Ba(OH)2.

Number of O atoms in 0.20 moles Ba(OH)2 = 2*0.20 = 0.40 moles O atom

In H2SO4 there are 4 moles of O atoms for every mol of H2SO4.

Number of O atoms in 0.20 moles H2SO4 = 4*0.20 = 0.80 moles O atom

⇒ 0.20 mol Ba(OH)2 and 0.20 mol H2SO4 do <u>not</u> have the same number of oxygen atoms.

<em>0.20 mol Br2O and 0.20 mol HBrO</em>

In Br2O there is 1 mol of O atoms in every mol Br2O

Number of O atoms in 0.20 moles Br2O = 0.20*1 = 0.20 moles O atom

In HBrO there is 1 mol of O atom in every mol HBrO

Number of O atoms in 0.20 moles HBrO = 0.20 *1 = 0.20 moles O atom

⇒ in 0.20 moles Br2O and 0.20 moles HBrO we <u>have the same</u> number of oxygen atoms

0.20 moles of Oygen contains: = 0.20 * 6.022*10^23 = 1.2 *10^23 O atoms

<em>0.10 mol Fe2O3 and 0.50 mol BaO</em>

In Fe2O3 there are 3 moles of O atoms in every mol of Fe2O3

Number of O atoms in 0.10 moles Fe2O3 = 0.10 * 3 = 0.30 moles O atom

In BaO there is 1 mol of O atoms in every mol BaO

Number of O atoms in 0.50 mol BaO = 1*0.50 = 0.50 moles O atom

⇒ 0.10 mol Fe2O3 and 0.50 mol BaO do <u>not</u> have the same number of oxygen atoms.

<em>0.10 mol Na2O and 0.10 mol Na2SO4</em>

In Na2O there is 1 mol of O atoms in every mol of Na2O

Number of O atoms in 0.10 mol Na2O = 1*0.10 = 0.10 moles O atom

In Na2sO4 there are 4 moles of O atoms in every mol of Na2SO4

Number of O atoms in 0.10 mol Na2SO4 = 4*0.10 = 0.40 moles O atom

⇒ 0.10 mol Na2O and 0.10 mol Na2SO4 do <u>not</u> have the same number of oxygen atoms.

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175 mL of Cl2 gas is held in a flexible vessel at
Gnoma [55]

Answer:

V₂ = 15.6 L

Explanation:

Given data:

Initial volume = 175 mL  (0.175 L)

Initial pressure = 1 atm

Initial temperature = 273 K

Final temperature = -5°C (-5+273 = 268 K)

Final volume = ?

Final pressure = 1.16 kpa (1.16/101=0.011 atm)

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm

V₂ = 46.9 L / 3.003

V₂ = 15.6 L

7 0
1 year ago
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What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
german
When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

PV = nRT

when P is the pressure =  1.1 atm

and V is the volume = 14.5 L 

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

1.1 * 14.5 L = n * 0.0821 * 285.5

∴ n = 1.1 * 14.5 / (0.0821*285.5)

       = 0.68 moles SO2

∴ moles CaSO3 = 0.68 moles

so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol


∴ mass = 0.68 moles* 120 g/mol = 81.6 g
7 0
2 years ago
The temperature of the carbon dioxide atmosphere near the surface of venus is 475°c. calculate the average kinetic energy per mo
harkovskaia [24]
Answer: <span>9330 j/mol
</span>
The temperature of the gas is 475 ° Celcius which is equal to: 475 +273= 748 °K. The formula for kinetic energy of individual atoms would be

K= 3/2 * kB * T
If kB is 1.38 * 10^-23 J/K and 1 mol is made from 6.02*10^23 molecule, then the kinetic energy of 1 mol CO2 would be:
K= 3/2 * kB * T
K= 3/2 * 1.38 * 10^-23  * 748 * 6.02 *10^23 =9324 J/mol
8 0
1 year ago
Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
Mekhanik [1.2K]

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

                Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

5 0
1 year ago
Read 2 more answers
Stabiliţi numerele de oxidare ale tuturor elementelor prezente în următoarele substanţe chimice, ţinând cont de principalele reg
natita [175]

Answer:

a.

N = +2

O = -2

b.

Na = +1

O = -2

N = +5

c.

O = -2

Al = +3

P = +5

d.

O = -2

Ca = +2

C = +4

e

O = -2

Mn = +7

K = +1

f.

O = -2

K = +1

Cr = +6

g.

O = -2

Cl = +7

Cr = +1

h.

O = -2

H = +1

Cr = +5

i.

O = -2

H = +1

Cr = +3

k.

O = -2

H = +1

Cr = +1

Explanation:

A. NU

Numărul de oxidare a azotului = +2

Numărul de oxidare a oxigenului = -2

b. NaNO₃

Numărul de oxidare de Na = +1

Numărul de oxidare de O = -2

Numărul de oxidare de N = +5

c. AlPO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Al = +3

Numărul de oxidare al P = +5

d. carbonat de calciu

Numărul de oxidare de O = -2

Numărul de oxidare de Ca = +2

Numărul de oxidare de C = +4

e. KMnO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Mn = +7

Numărul de oxidare al lui K = +1

f. K₂Cr₂O₇

Numărul de oxidare de O = -2

Numărul de oxidare al lui K = +1

Numărul de oxidare al Cr = +6

g. HClO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Cl = +7

Numărul de oxidare al Cr = +1

h. HClO₃

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +5

i. HClO₂

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +3

k. HClO

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +1.

6 0
1 year ago
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